MOOCULUS-2 "Sequences and Series" 数列与级数 学习笔记: 1. Sequences

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此课程由Ohio University于2014年在Coursera平台讲授。

本系列学习笔记PDF下载（Academia.edu） MOOCULUS-2 Solution

Summary

• Suppose that $\left(a_n\right)$ is a sequence. To say that $\lim_{n\to \infty}a_n=L$ is to say that for every $\varepsilon>0$, there is an $N > 0$, so that whenever $n>N$, we have $|a_n-L| < \varepsilon$. If $\lim_{n\to\infty}a_n=L$ we say that the sequence converges. If there is no finite value $L$ so that $\lim_{n\to\infty}a_n = L$, then we say that the limit does not exist, or equivalently that the sequence diverges.
• Suppose $(a_n)$ is a sequence with initial index $N$, and suppose we have a sequence of integers $(n_i)$ so that $$N \leq n_1 < n_2 < n_3 < n_4 < n_5 < \cdots$$ Then the sequence $(b_i)$ given by $b_i = a_{n_i}$ is said to be a subsequence of the sequence $a_n$.
• If $(b_i)$ is a subsequence of the convergent sequence $(a_n)$, then $$\lim_{i \to \infty} b_i = \lim_{n \to \infty} a_n$$
• Suppose $(b_i)$ and $(c_i)$ are convergent subsequences of the sequence $(a_n)$, but $$\lim_{i \to \infty} b_i \neq \lim_{i \to \infty} c_i.$$ Then the sequence $(a_n)$ does not converge.
• Squeeze Theorem: Suppose there is some $N$ so that for all $n > N$, it is the case that $a_n \le b_n \le c_n$. If $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}c_n=L$$ then $\lim_{n\to\infty}b_n=L$.
• $$\lim_{n\to\infty}|a_n|=0$$ if and only if $$\lim_{n\to\infty}a_n=0$$
• The sequence $a_n = r^n$ converges when $-1 < r \le 1$, and diverges otherwise. In symbols, $$\lim_{n\to\infty} r^n=\begin{cases}0& \mbox{if $-1 < r < 1$,} \\ 1& \mbox{if $r=1$, and} \\ \mbox{does not exist} & \mbox{if $r \leq -1$ or $r > 1$.} \end{cases}$$
• A sequence is called increasing (or sometimes strictly increasing) if $a_n < a_{n+1}$ for all $n$. It is called non-decreasing if $a_n\le a_{n+1}$ for all $n$. Similarly a sequence is decreasing (or, by some people, strictly decreasing) if $a_n > a_{n+1}$ for all $n$ and non-increasing if $a_n\ge a_{n+1}$ for all $n$.
• If a sequence is increasing, non-decreasing, decreasing, or non-increasing, it is said to be monotonic.
• A sequence $(a_n)$ is bounded above if there is some number $M$ so that for all $n$, we have $a_n\le M$. Likewise, a sequence $(a_n)$ is bounded below if there is some number $M$ so that for every $n$, we have $a_n\ge M$. If a sequence is both bounded above and bounded below, the sequence is said to be bounded.
• If the sequence $a_n$ is bounded and monotonic, then $\lim_{n \to \infty} a_n$ exists. In short, bounded monotonic sequences converge.

Exercises

1. Compute $$\lim_{x\to\infty} x^{1/x}$$ Solution: $$\lim_{x\to\infty} x^{1/x}=\lim_{x\to\infty}(e^{\ln x})^{1/x} =\lim_{x\to\infty}e^{\frac{\ln x}{x}}$$ By L‘Hopital‘s rule, we have $$\lim_{x\to\infty}\frac{\ln x}{x}=\lim_{x\to\infty}\frac{1/x}{1}=0$$ Thus, the result is $$\lim_{x\to\infty} x^{1/x}=e^0=1$$ 2. Use the squeeze theorem to show that $$\lim_{n\to\infty} {n!\over n^n}=0$$ Solution: $$0<\frac{n!}{n^n}=\frac{1}{n}\cdot\frac{2}{n}\cdot\cdots\cdots\cdot\frac{n}{n} < \frac{1}{n}\to0\ (n\to\infty)$$ According to the squeeze theorem, $$\lim_{n\to\infty} {n!\over n^n}=0$$ 3. Determine whether $$\{\sqrt{n+47}-\sqrt{n}\}_{n=0}^\infty$$ converges or diverges. If it converges, compute the limit.

Solution: $$\sqrt{n+47}-\sqrt{n}=\frac{47}{\sqrt{n+47}+\sqrt{n}}$$ Hence it is decreasing. On the other hand, $$\sqrt{n+47}-\sqrt{n}\ge0$$ that is, it is bounded below. Thus, it is convergent. And we have $$\lim_{x\to\infty}(\sqrt{n+47}-\sqrt{n})=\lim_{n\to\infty}\frac{47}{\sqrt{n+47}+\sqrt{n}}=0$$ 4. Determine whether $$\left\{{n^2+1\over (n+1)^2}\right\}_{n=0}^\infty$$ converges or diverges. If it converges, compute the limit.

Solution: $${(n+1)^2\over n^2+1}=1+{2n\over n^2+1}$$ which is decreasing. Thus $${n^2+1\over (n+1)^2}$$ is increasing. On the other hand, $${n^2+1\over (n+1)^2}={n^2+1\over n^2+2n+1} < 1$$ which means it is bounded above. Thus it is convergent. And we have $$\lim_{n\to\infty}{n^2+1\over (n+1)^2}=\lim_{n\to\infty}\frac{n^2+1}{n^2+2n+1}=\lim_{n\to\infty}\frac{1+\frac{1}{n^2}}{1+\frac{2}{n}+\frac{1}{n^2}}=1$$ 5. Determine whether $$\left\{{n+47\over\sqrt{n^2+3n}}\right\}_{n=1}^\infty$$ converges or diverges. If it converges, compute the limit.

Solution: $$f^{‘}(n)=\frac{\sqrt{n^2+3n}-(n+47)\cdot{1\over2}\cdot{1\over\sqrt{n^2+3n}}\cdot(2n+3)}{n^2+3n} < 0$$ $$\Longleftrightarrow \sqrt{n^2+3n}-(n+47)\cdot{1\over2}\cdot{1\over\sqrt{n^2+3n}}\cdot(2n+3) < 0$$ $$\Longleftrightarrow n^2+3n < {1\over2}\cdot(2n^2+97n+141)$$ $$\Longleftrightarrow n^2+3n < n^2+48.5n+70.5$$ The last inequality is obvious. Thus it is decreasing. On the other hand, $${n+47\over\sqrt{n^2+3n}}>0$$ which means it is bounded below. Hence it is convergent. And we have $$\lim_{n\to\infty}{n+47\over\sqrt{n^2+3n}}=\lim_{n\to\infty}{1+\frac{47}{n}\over\sqrt{1+\frac{3}{n}}}=1$$ 6. Determine whether $$\left\{{2^n\over n!}\right\}_{n=0}^\infty$$ converges or diverges. If it converges, compute the limit.

Solution: $${a_{n+1}\over a_n}={\frac{2^{n+1}}{(n+1)!}\over\frac{2^n}{n!}}={2\over n+1} < 1$$ when $n > 2$. Thus it is decreasing. On the other hand, $${2^n\over n!}>0$$ which means it is bounded below. Thus it is convergent. $$0<{2^n\over n!}={2\over n}\cdot {2\over n-1} \cdot\cdots\cdots\cdot{2\over3}\cdot{2\over2}\cdot{2\over1} < ({2\over3})^{n-2}\cdot2\to0\ (n\to\infty)$$ According to squeeze theorem we have $$\lim_{n\to\infty}{2^n\over n!}=0$$

MOOCULUS-2 "Sequences and Series" 数列与级数 学习笔记: 1. Sequences

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原文地址：http://www.cnblogs.com/zhaoyin/p/4121880.html